# Prove that the average of $D_wD_wf(x_0,y_0)$ over all unit vectors $w$ is equal to $\frac{1}{2} \Delta f(x_0,y_0)$ for any smooth function $f$.

Here is a challenge problem from my math professor:

Let $w$ be a unit vector in $\mathbb{R}^2$, and let $D_w$ denote the directional derivative with respect to $w$. Prove that for any smooth function $f$ on $\mathbb{R}^2$ and any point $(x_0, y_0) \in \mathbb{R}^2$, the average of $D_w D_w f(x_0, y_0)$ over all unit vectors $w$ is equal to $\frac{1}{2}\Delta f(x_0, y_0)$. This fact (which generalizes to higher dimensions) helps to explain why the Laplacian $\Delta$ is a "natural" differential operator that appears in many important PDEs.

My Attempt

Assuming that $f$ is any smooth function, we see that their partial derivatives exist. For my workout, I set the unit vector to be $w = (a,b)$. I tried to prove this statement via computation as follows:

For the left-hand side...

\begin{aligned} D_wD_w f(x,y) &= D_w((a,b) \cdot (f_x, f_y))\\ &= D_w(af_x + bf_y)\\ &= (a,b) \cdot (af_{xx} + bf_{yx}, af_{xy} + bf_{yy})\\ &= a^2f_{xx} + 2abf_{xy} + b^2f_{yy}\\ &= a^2f_{xx} + b^2f_{yy} \end{aligned}

For the right-hand side...

\begin{aligned} \dfrac{1}{2}\Delta f(x,y) = \dfrac{1}{2}(f_{xx} + f_{yy}) \end{aligned}

So at the point $(x_0, y_0) \in \mathbb{R}^2$,

\begin{aligned} D_wD_w f(x_0,y_0) &= a^2f_{xx}(x_0, y_0) + b^2f_{yy}(x_0, y_0)\\ \dfrac{1}{2}\Delta f(x_0,y_0) &= \dfrac{1}{2}(f_{xx}(x_0, y_0) + f_{yy}(x_0, y_0)) \end{aligned}

I'm stuck here since I am not sure how to approach this problem because of "average of directional derivatives over the unit vectors".

All unit vectors can be parametrized by $(a,b)=(\cos\theta,\sin\theta)$, so you need to average $D_wD_wf(x_0,y_0)$ over the circle, and since $\frac{1}{2\pi}\int_0^{2\pi}\cos^2\theta\,d\theta=\frac{1}{2\pi}\int_0^{2\pi}\sin^2\theta\,d\theta=\frac{1}{2}$, we get the result.

By the way, you made an error in $D_wD_wf(x_0,y_0)$, you for some reason dropped the $f_{xy}$ term. This term should be there, however it will disappear when you do the average.

• Very sweet answer! Thanks! – NasuSama Feb 1 '14 at 21:45
• @NasuSama No problem. Also I edited my answer because I noticed you made a mistake in the computation of $D_wD_wf(x_0,y_0)$. – JLA Feb 1 '14 at 21:48

Our OP NasuSama has done a lot of the work already! We have, as the fruit of his efforts,

$D_wD_w f(x, y) = a^2f_{xx} + 2abf_{xy} +b^2f_{yy}; \tag{1}$

however, it is not in general true that

$a^2f_{xx} + 2abf_{xy} + b^2f_{yy} = a^2f_{xx} + b^2f_{yy}, \tag{2}$

since this holds if and only if

$2abf_{xy} = 0, \tag{3}$

which is of course false unless $f_{xy} = 0$ or $ab= 0$, which for unit vectors $(a, b)$ means $a = \pm 1, b = 0$ or $a= 0, b = \pm 1$ since $a^2 + b^2 = 1$. If, on the other hand, we take the average of (1) over all unit vectors $w= (a, b)$ at some point $(x_0, y_0)$, the term $2f_{xy}(x_0, y_0)ab$ washes out, thus: unit vectors $w = (a, b)$ may be represented as $(a, b) = (\cos \theta, \sin \theta)$, where $\theta \in [0, 2\pi)$. Bearing this in mind, (1) becomes, at the point $(x_0, y_0)$,

$D_wD_wf(x_0, y_0) = \cos^2 \theta f_{xx}(x_0, y_0) + 2\cos \theta \sin \theta f_{xy}(x_0, y_0) + \sin^2 \theta f_{yy}(x_0, y_0); \tag{4}$

we average (4) over unit vectors by applying the integral operator $(1 / 2 \pi) \int_0^{2\pi} d \theta$ to each side, obtaining, by virtue of the fact that the derivatives of $f$ are independent of the variable $\theta$,

$\dfrac{1}{2\pi}\int_0^{2\pi}(D_wD_w)f(x_0, y_0) d\theta$ $= f_{xx}(x_0, y_0)\dfrac{1}{2\pi}\int_0^{2\pi}\cos^2 \theta d\theta + 2 f_{xy}(x_0, y_0)\dfrac{1}{2\pi}\int_0^{2\pi}\cos \theta \sin \theta d\theta$ $+ f_{yy}(x_0, y_0)\dfrac{1}{2\pi}\int_0^{2\pi}\sin^2 \theta d\theta. \tag{5}$

Since $\int_0^{2\pi} \cos^2 \theta d \theta = \int_0^{2\pi} \sin^2 \theta d \theta = \pi$ and $\int_0^{2\pi}\cos \theta \sin \theta d\theta = (\dfrac{1}{2}\sin^2 \theta \mid_0^{2\pi} = 0$,

(5) becomes

$\dfrac{1}{2\pi}\int_0^{2\pi}(D_wD_w)f(x_0, y_0) d\theta = \dfrac{1}{2}(f_{xx}(x_0, y_0) + f_{yy}(x_0, y_0)), \tag{6}$

the requisite result. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!